package org.flint.summary.leetcode;

/**
 * @author flint92
 */
public class CountingBits_338 {

    /**
     * Given a non negative integer number num.
     * For every numbers i in the range 0 ≤ i ≤ num
     * calculate the number of 1's in their binary representation and return them as an array.
     * <p>
     * Example 1:
     * <p>
     * Input: 2
     * Output: [0,1,1]
     * Example 2:
     * <p>
     * Input: 5
     * Output: [0,1,1,2,1,2]
     * Follow up:
     * <p>
     * It is very easy to come up with a solution with run time O(n*sizeof(integer)).
     * But can you do it in linear time O(n) /possibly in a single pass?
     * Space complexity should be O(n).
     * Can you do it like a boss?
     * Do it without using any builtin function like __builtin_popcount in c++ or in any other language.
     */
    public int[] countBits(int num) {

        int[] bits = new int[num + 1];
        for (int i = 1; i <= num; i++) {
            // i & (i - 1)表示去掉i的二进制数最低位的1
            // 由上可知 i > (i & (i - 1))
            // 而bits[i & (i - 1)]肯定在之前的循环中已经求出
            // 则直接求出bits[i] = bits[i & (i - 1)] + 1;
            bits[i] = bits[i & (i - 1)] + 1;
        }
        return bits;
    }
}
